Since this is rotated around the y axis, using shells, you should be integrating with respect to x, not y.

From , with x between 2-1= 1 and 2+1= 3. Each shell is a cylinder with height and radius x so the volume is given by 2\pi\int_{x=1}^3 x\sqrt{-5+ 4x- x^2}dxp[/tex]

I think I would be inclined, rather, to use "disks" or "washers". From , for each y. That is, a cross section of the solid at each y is a "washer" lying between the two circles of radius and . You can calculate the area of the "washer" between the two circles as the difference between the area of the two circles: and the volume is

2pi [(1/3)(sqrt(1-(x-2)^2))(x^2-x-3) - arcsin(2-x)] from x=1 to x=3, I ended up getting pi inside the square brackets which resulted in the answer being 3pi with the 2pi on the outside, so maybe just trying subbing x=1 and x=3 again.

But my question to you is, how did you integrate:

2pi[ int(x(sqrt (1-(x-2)^2))) ] from x=1 to x=3 into the integral you got above? I'm curious because I couldn't do it myself